The probability of selecting red in the 1st selection would be $\frac 1 7$. Given this, and since replacement is allowed, in the second selection, the probability would be $\frac 1 7 \times \frac 1 7 = \frac {1} {49} $
The previous answer is correct, if there is exactly one red object among seven. The other colors are meaningless, they all may be even the same. If there are k red objects and 7-k are non-red, then the answer is k2/49.