= Solution
{c}
At the <Köhler activation radius>, the two coefficients satisfy
$$
S_c-1=\frac{2A}{3r_c},\qquad
A=\frac32r_c(S_c-1).
$$
Converting the <atmospheric supersaturation> to a fraction gives $S_c-1=0.004$, not $0.4$. Hence
$$
A=\frac32(2.0\times10^{-7}\,\mathrm m)(0.004)
=1.2\times10^{-9}\,\mathrm m,
$$
and the supplied curvature coefficient gives the <temperature>
$$
\boxed{T=\frac{3.3\times10^{-7}\,\mathrm{K\,m}}{1.2\times10^{-9}\,\mathrm m}
=275\,\mathrm K\simeq1.9\,{}^\circ\mathrm C.}
$$
The coefficient and the stated radius consistently use a radius convention; replacing radius by diameter without changing the coefficient would introduce a factor-of-two error.
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