Answer
$(P_{\rm{tunnel}})_d \gt (P_{\rm{tunnel}})_a \gt (P_{\rm{tunnel}})_b \gt (P_{\rm{tunnel}})_c$
Work Step by Step
We know that
$$P_{\rm{tunnel}} = e^{\frac{-2w}{\eta}}$$
where $\eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}}$
So,
$$P_{\rm {tunnel}} = e^{\left[\dfrac{-2w\sqrt{2m(U_0 - E)}}{\hbar}\right]}$$
Applying this formula to the 4 barriers.
$\textbf{For barrier a:}$
$$(P_{\rm {tunnel}})_a = e^{\left[\dfrac{-2w\sqrt{2m(1)}}{\hbar}\right]}$$
$$\boxed{(P_{\rm {tunnel}})_a = e^{\left[\dfrac{-2.83\;w\sqrt{ m }}{\hbar}\right]}}$$
$\textbf{For barrier b:}$
$$(P_{\rm {tunnel}})_b = e^{\left[\dfrac{-2w\sqrt{2m(2)}}{\hbar}\right]}$$
$$\boxed{(P_{\rm {tunnel}})_b= e^{\left[\dfrac{-4\;w\sqrt{ m }}{\hbar}\right]}}$$
$\textbf{For barrier c:}$
$$(P_{\rm {tunnel}})_c = e^{\left[\dfrac{-2(2w)\sqrt{2m(1)}}{\hbar}\right]}$$
$$\boxed{(P_{\rm {tunnel}})_c= e^{\left[\dfrac{-5.66\;w\sqrt{ m }}{\hbar}\right]}}$$
$\textbf{For barrier d:}$
$$(P_{\rm {tunnel}})_d = e^{\left[\dfrac{-2(0.5w)\sqrt{2m(2)}}{\hbar}\right]}$$
$$\boxed{ (P_{\rm {tunnel}})_d= e^{\left[\dfrac{-2\;w\sqrt{ m }}{\hbar}\right]}}$$
We know that the larger the negative exponent, the smaller the value.
So that the ranking is then as follows:
$$
\boxed{\boxed{(P_{\rm{tunnel}})_d \gt (P_{\rm{tunnel}})_a \gt (P_{\rm{tunnel}})_b \gt (P_{\rm{tunnel}})_c}}\;
$$
