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Let $r$ be the radius of the sphere. The volume $V$ of the sphere is given by $V = \frac{4}{3}\pi r^3$. The surface area $S$ of the sphere is given by $S = 4\pi r^2$. We are given that the rate of change of volume with respect to time is equal to the rate of change of radius with respect to time, i.e., $\frac{dV}{dt} = \frac{dr}{dt}$.
Using the chain rule, differentiate $V$ with respect to $t$: $$ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi r^3 \right) = 4\pi r^2 \frac{dr}{dt} $$
Substitute the given condition $\frac{dV}{dt} = \frac{dr}{dt}$ into the equation: $$ \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt} $$ Assuming $\frac{dr}{dt} \neq 0$, we divide both sides by $\frac{dr}{dt}$: $$ 1 = 4\pi r^2 $$
The surface area $S$ is $4\pi r^2$. From the equation derived in Step 3, we already have $4\pi r^2 = 1$. Therefore, the surface area is 1 sq. unit.
Final Answer: 1 sq. unit