= Solution
Interpret the quoted r.m.s. as the root-mean-square fluctuation about the <mean>, namely the <standard deviation> $\sigma=10\,\mathrm K$. Standardization gives $z=(293-273)/10=2$. The supplied cumulative Gaussian curve gives $\Phi(2)\simeq0.977$, hence
$$
\boxed{P(T>293\,\mathrm K)=1-\Phi(2)\simeq0.023.}
$$
The exact Gaussian value is $0.02275$. A raw r.m.s. <temperature> of $10\,\mathrm K$ could not coexist with a <mean> of $273\,\mathrm K$; the spread interpretation is required.
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