Answer
$\eta_a = \eta_b\gt \eta_c$
Work Step by Step
We know that the penetration distance is given by
$$\eta=\dfrac{\hbar}{\sqrt{2m(U_0-E)}}$$
So the larger the difference $ U_0 - E $, the smaller the penetration distance.
$\underline{\text{Analyzing Each Case}}$
$\underline{\text{Figure (a):}}$
$ U_0 = 10 \; \rm {eV} $, $ E = 5 \; \rm {eV} $, so the difference $U_0 - E = 10 - 5 = 5 \; \rm {eV} $.
Therefore, $$ \eta_a = \dfrac{\hbar}{\sqrt{10\;m}}\tag 1 $$
$\underline{\text{Figure (b):}}$
$ U_0 = 10 \; \rm {eV} $, $ E = 5 \; \rm {eV} $, so the difference $U_0 - E = 10 - 5 = 5 \; \rm {eV} $.
Therefore, $$ \eta_b = \dfrac{\hbar}{\sqrt{10\;m}} \tag 2$$
Thus, from (1) and (2),
$$\eta_a=\eta_b$$
$\underline{\text{Figure (c):}}$
$ U_0 = 16\; \rm {eV} $, $ E = 10\; \rm {eV} $, so the difference $U_0 - E = 16- 10= 6\; \rm {eV} $.
Therefore, $$ \eta_c = \dfrac{\hbar}{\sqrt{12\;m}}\tag 3 $$
From (1), (2), (3);
$$ \boxed{\eta_a = \eta_b\gt \eta_c} $$
