You can solve it using Convolution Theorem:

Fourier transform of a convolution is the point wise product of Fourier transforms.

Since usual Fourier transform of the function t is not exist you should use the theory of distributions. See the book of Loran Schwartz or Lars Hormander.

Use The Sumudu transform. You can find similar ideas in this paper" 

https://www.researchgate.net/publication/26405886_Analytical_investigations_of_the_Sumudu_transform_and_applications_to_integral_production_equations

You can also check the following papers to know more about this poweful transform

https://www.researchgate.net/publication/41448911_Sumudu_transform_fundamental_properties_investigations_and_applications

https://www.researchgate.net/publication/249008344_Introducing_and_Analysing_Deeper_Sumudu_Properties

https://www.researchgate.net/publication/233420835_Sumudu_Transform_Applications_to_Bessel_Functions_and_Equations

Article Analytical investigations of the Sumudu transform and applic...

Article Sumudu transform fundamental properties investigations and a...

Article Introducing and Analysing Deeper Sumudu Properties

Article Sumudu Transform Applications to Bessel Functions and Equations

Dear Muhammad,

I am sorry for my late answer, since I noticed your question just a moment ago. I would say that in general it is impossible to solve the equation explicitly. Also, the answer depends on the notion of the convolution. Is it the usual convolution over R or the Laplace convolution over the interval (0,t).

If the convolution is over R, you could in principle use the Fourier transform as it was suggested in the first comment but still you have to work with the uncomfortable Dirac delta distribution and its derivatives together with the convolution (since there is a multiplication with a(t)) on the Fourier side also. Then, if you transform back the resulting equation, you will end up to even a more complicated equation with integrals (if everything went ok in my reasoning).

On the other hand, if the convolution means the Laplace convolution, you can derive the corresponding differential equation. But I would say that the differential equation can be solved explicitly only for special n's and a(t)'s. Indeed, if n=0, then dividing the both sides by a(t) you get an equation 1/a(t)u(t)+\int_0^t u(s)ds=t/a(t). If you differentiate this equation once, you will get a first order ODE, which can be solved explicitly (provided a(t) is differentiable and non-zero everywhere). If n>0, then the equation is of the form

(1) 1/a(t)u(t)+n! (J^{n+1} u)(t)=t/a(t),

where J^{n+1}=1/n! \int_0^t (t-s)^n u(s)ds is the iterated integral of u of order (n+1). Then if u and a are smooth enough, you can differentiate the equation (1) (n+1) times to get an ODE of order (n+1). But this can be solved explicitly only for special cases of the function a(t) (e.g. the case 'a(t) is a constant' is fine).  

Hopefully my answer helped you a little bit.