1. Given centripetal acceleration

For circular motion of radius $r$, the centripetal acceleration is

$a_c = \frac{v^2}{r}$

The problem gives

$a_c = k^2 r t^2$

So,

$\frac{v^2}{r} = k^2 r t^2$

$v^2 = k^2 r^2 t^2$

Taking the positive root for speed,

$v = krt$

2. Find tangential acceleration

Power delivered by the force is due to the component of force along velocity, i.e. the tangential force.

Tangential acceleration is

$a_t = \frac{dv}{dt} = \frac{d}{dt}(krt) = kr$

Thus tangential force is

$F_t = ma_t = mkr$

3. Compute power

Instantaneous power is

$P = \vec F \cdot \vec v = F_t \, v$

Substituting,

$P = (mkr)(krt)$

$P = mk^2 r^2 t$

4. Check options

• A: $0$ ❌
• B: $mk^2 r^2 t^2$ ❌
• C: $mk^2 r^2 t$ ✅
• D: $mk^2 r t$ ❌

5. Final answer

$\boxed{P = mk^2 r^2 t}$

So, the correct option is C.