Step 1: Find the temperature ratio:
A peak at shorter wavelength means a hotter body. If $\lambda_m \to \frac23\lambda_m$, then $T \to \frac32T$.
Step 2: Apply Stefan's law:
$E\propto T^4$, so $E' = E\times\left(\frac32\right)^4 = E\times\frac{3^4}{2^4}$.
$3^4 = 81$ and $2^4 = 16$.
Step 3: Result:
$E' = \frac{81}{16}E$, option (B). A common mistake is to use $T^2$, which gives $\frac94E = \frac{36}{16}E$, a value not in the list.
Final Answer:
Option (B).
\[ \boxed{\frac{81}{16}E \text{ (B)}} \]