Given expressions

$V=100\sin(100t)\ \text{V}$ $I=100\sin\left(100t+\frac{\pi}{3}\right)\ \text{mA}$

Convert current to ampere: $I=0.1\sin\left(100t+\frac{\pi}{3}\right)\ \text{A}$

Identify peak values and phase difference

From the equations: $V_0=100\ \text{V},\qquad I_0=0.1\ \text{A}$

Phase difference: $\phi=\frac{\pi}{3}$

Formula for average power in AC

The average power over one cycle is: $P_{\text{avg}}=V_{\text{rms}}I_{\text{rms}}\cos\phi$

where $V_{\text{rms}}=\frac{V_0}{\sqrt{2}}=\frac{100}{\sqrt{2}}$ $I_{\text{rms}}=\frac{I_0}{\sqrt{2}}=\frac{0.1}{\sqrt{2}}$

Substitute values

$P_{\text{avg}}=\left(\frac{100}{\sqrt{2}}\right)\left(\frac{0.1}{\sqrt{2}}\right)\cos\frac{\pi}{3}$

Since $\cos\frac{\pi}{3}=\frac{1}{2}$

and $\left(\frac{100}{\sqrt{2}}\right)\left(\frac{0.1}{\sqrt{2}}\right)=\frac{10}{2}=5$

so, $P_{\text{avg}}=5\times \frac{1}{2}=2.5\ \text{W}$

Check with options

$2.5\ \text{W}$ corresponds to Option C.

Comparison with stored answer

Stored correct answer: C

Derived answer: C

So they agree.