= Solution
The radius is half the diameter, $r^*=1.5\times10^{-9}\ \mathrm m$, and the <latent heat> density is $1.25\times10^9\ \mathrm{J\,m^{-3}}$. Hence
$$
\Delta T=\frac{2(0.3)(1337)}{(1.5\times10^{-9})(1.25\times10^9)}=427.84\ \mathrm K,
\qquad\boxed{T_m(3\ \mathrm{nm})\simeq909\ \mathrm K.}
$$
Using the diameter directly in the radius formula would underestimate the depression by a factor of two. The result is the stipulated continuum-model estimate; at only a few nanometres atomistic structure and support effects can matter substantially.
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