= Solution
For the liquid pump take the explicit incompressible-water estimate $v_f\simeq10^{-3}\,\mathrm{m^3\,kg^{-1}}$. Its work input is
$$
w_p\simeq v_f(P_2-P_1)=10^{-3}(28-0.002)10^6=27.998\,\mathrm{kJ\,kg^{-1}},
$$
so $h_2=h_1+w_p=111.998\,\mathrm{kJ\,kg^{-1}}$. With $h_3=3463$ and the preceding $h_4=1838.74$ in the same units,
$$
q_{\rm in}=h_3-h_2=3351.002,\qquad
w_{\rm net}=h_3-h_4-w_p=1596.26.
$$
Thus
$$
\boxed{\eta=\frac{w_{\rm net}}{q_{\rm in}}=0.47635\simeq47.6\%.}
$$
If pump work is neglected because no liquid density was tabulated, the corresponding approximation is $(3463-1838.74)/(3463-84)=48.1\%$. Stating the approximation explains the small difference. The temperature-extreme <Carnot efficiency> is about $1-293.15/873.15=66.4\%$, above both results; finite-temperature-range heat addition in the <Rankine cycle> does not attain that limit.
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