According to the question, room-glass interface and glass-outdoor interface are connected in series combination so, the equivalent thermal resistance,
$$
\begin{aligned}
& R_{e q}=R_1+R_2+R_3 \\
\because & R_1=\frac{l_1}{k_1 A_1}, R_2=\frac{l_2}{k_2 A_2}, R_3=\frac{l_3}{k_3 A_3}
\end{aligned}
$$

Where, $k_1$ and $k_3$ are thermal conductivity of glass and $k_2$ is thermal conductivity of air.
or
$$
R_{e q}=\frac{l_1}{k_1 A_1}+\frac{l_2}{k_2 A_2}+\frac{l_3}{k_3 A_3}
$$

Given, $l_1=l_3=1 \mathrm{~cm}=10^{-2} \mathrm{~m}$,
$$
\begin{aligned}
l_2 & =5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m} \\
k_1 & =k_3=0.8 \mathrm{Wm}^{-1} \mathrm{~K}^{-1}, \\
k_2 & =0.08 \mathrm{Wm}^{-1} \mathrm{~K}^{-1} \\
\text { and } A_1=A_2 & =A_3=2.6 \mathrm{~m}^2
\end{aligned}
$$

Putting the given values, we get
$$
\begin{aligned}
& \frac{10^{-2}}{0.8 \times 2.6}+\frac{5 \times 10^{-2}}{0.08 \times 2.6}+\frac{10^{-2}}{0.8 \times 2.6} \\
& =\left(\frac{10^{-2}}{2.6}\right)\left(\frac{52}{0.8}\right)=\frac{1}{4}
\end{aligned}
$$

Hence, flow of heat,
$$
\begin{aligned}
& H=\frac{\Delta T}{R_{\text {eq }}}=\frac{[18-(-2)]}{\left(\frac{1}{4}\right)} \\
& H=80 \mathrm{~W}
\end{aligned}
$$