= Solution
The <Clapeyron equation> for molar <latent heat> $L$ is $dp/dT=L/[T(V_{\rm vap}-V_{\rm liq})]$. Neglect the liquid <molar volume>, treat the vapor as an <ideal gas>, and approximate $L$ as constant over the interval. Then
$$
\frac{d\ln p}{dT}\simeq\frac{L}{RT^2},\qquad
\ln\frac{p_h}{p_l}=\frac LR\left(\frac1{T_l}-\frac1{T_h}\right).
$$
Using one atmosphere as $101325\,\mathrm{Pa}$ gives
$$
\boxed{L\simeq\frac{R\ln(101325/133)}{1/1.2-1/4.2}=92.7\,\mathrm{J\,mol^{-1}}.}
$$
This is the effective molar <enthalpy of vaporization> inferred from the <Integrated Clausius-Clapeyron relation>. If $L$ varies with <temperature>, the inferred value is weighted by $1/T^2$, not an unweighted arithmetic mean. Real helium is not exactly an <ideal gas> over the whole interval, and its liquid changes superfluid state within it, so this is an estimate under the stated approximations rather than a precise temperature-independent material constant.
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