$$
begin{align*}
P_{avg}&=I^2_{rms}cdot{R}\
P_{avg}&=left(frac{V_{max}}{sqrt{2}cdot{R}}right)^2cdot{R}\
P_{avg}&=frac{left(V_{max}right)^2}{2cdot{R}}
end{align*}
$$

So, to have dissipation of $P_{avg}=5text{ W}$ the resistor resistance have to be:

$$
begin{align*}
R&=frac{left(V_{max}right)^2}{2cdot{P_{avg}}}\
R&=frac{left(24text{ V}right)^2}{2cdot{5text{ W}}}\
R&=57.6text{ $Omega$}
end{align*}
$$

As this value is lower then the value that we get at the example $23.8$,

$$
boxed{text{The resistance have to be decreased to $R=57.6text{ $Omega$}$.}}
$$

$$
begin{align*}
P_{avg}&=I^2_{rms}cdot{R}\
P_{avg}&=left(frac{V_{max}}{sqrt{2}cdot{R}}right)^2cdot{R}\
P_{avg}&=frac{left(V_{max}right)^2}{2cdot{R}}
end{align*}
$$

Let’s express the maximum voltage from this equation:

$$
begin{align*}
left(V_{max}right)^2&=2cdot{P_{avg}}cdot{R}\
V_{max}&=sqrt{2cdot{P_{avg}}cdot{R}}
end{align*}
$$

Let’s substitute and compute the value:

$$
begin{align*}
V_{max}&=sqrt{2cdot{2.25text{ W}}cdot{375text{ $Omega$}}}
end{align*}
$$

$$
boxed{V_{max}=41.08text{ V}}
$$

$$
begin{align*}
P_{avg}&=I^2_{rms}cdot{R}\
P_{avg}&=I^2_{max}cdotfrac{R}{2}
end{align*}
$$

Let’s express current from this equation:

$$
begin{align*}
I^2_{max}&=frac{2cdot{P_{avg}}}{R}\
I_{max}&=sqrt{frac{2cdot{P_{avg}}}{R}}\
I_{max}&=sqrt{frac{2cdot{4.45text{ W}}}{232text{ $Omega$}}}
end{align*}
$$

$$
boxed{I_{max}=0.2text{ A}}
$$