= Solution
The mean interface height fluctuation is zero; its <root mean square> roughness is obtained by summing the mode variances. Replacing the two-dimensional <wavevector> sum by its continuum density gives the <thermal roughness of a tension-dominated interface>:
$$
\langle h^2\rangle\simeq\frac{A}{(2\pi)^2}\int_{q_{\min}}^{q_{\max}}2\pi q\,dq\,\frac{k_BT}{A\gamma q^2}
=\frac{k_BT}{2\pi\gamma}\log\frac{q_{\max}}{q_{\min}}.
$$
Take $q_{\min}=2\pi/L$ with $L=\sqrt A=10^{-2}\ \mathrm m$, and $q_{\max}=2\pi/d$ with $d=2\times10^{-10}\ \mathrm m$. At $T=300\ \mathrm K$,
$$
\boxed{\langle h^2\rangle\simeq3.08\times10^{-19}\ \mathrm{m^2},\qquad\sqrt{\langle h^2\rangle}\simeq0.55\ \mathrm{nm}.}
$$
The logarithm makes the answer only weakly sensitive to order-one choices of cutoff. Gravity has been omitted as stipulated; the fixed mean height removes the zero <wavevector> mode.
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