Find the carrier power first, then scale it by the modulation factor.

Step 1 — the carrier power. The 150 V is a peak value, so the RMS value is \(150/\sqrt{2}\) and

\(P_{c}=\dfrac{V_{c}^{2}}{2R}=\dfrac{150^{2}}{2\times200}=\dfrac{22500}{400}=56.25\ \text{W}\)

The factor 2 in the denominator is exactly the squared \(\sqrt{2}\) of the peak-to-RMS conversion, and forgetting it doubles the answer — which is what option 4 nearly represents.

Step 2 — add the sideband power.

\(P_{t}=P_{c}\left(1+\dfrac{m^{2}}{2}\right)=56.25\left(1+\dfrac{0.5^{2}}{2}\right)\)

\(=56.25\left(1+0.125\right)=56.25\times1.125=63.28\ \text{W}\)

— option 3.

Where the m²/2 comes from. Each sideband has an amplitude of \(mV_{c}/2\), so each carries

\(P_{SB}=\dfrac{\left(mV_{c}/2\right)^{2}}{2R}=\dfrac{m^{2}}{4}P_{c}\)

and there are two of them, giving \(m^{2}P_{c}/2\) in total. Here that is 7.03 W of information-bearing power against 56.25 W of carrier.

What the numbers say about AM's efficiency. At \(m=0.5\) only 11 % of the transmitted power carries any information; the rest is spent on a carrier that conveys nothing. Even at full modulation, \(m=1\), the figure rises only to 33 %. That is the whole case for suppressed-carrier transmission — and the reason the carrier is nevertheless kept in broadcasting is that it allows a receiver to be built from a single diode.

A useful check on the result : the total can never be less than the carrier power alone, and for \(m\le1\) can never exceed \(1.5P_{c}=84.4\) W. Options 1 and 2 fall outside those bounds — 263.5 W is far too large and 45.23 W is below the carrier power — so both can be rejected before any arithmetic.

Hence, the total power is 63.28 W.