The probability of getting exactly 1 six when 6 dice are rolled
$= (_6C_1)(1/6)^1(5/6)^5$
The probability of getting exactly x sixes when 6 dice are rolled
$= (_6C_x)(1/6)^x (5/6)^{6-x} ~ ~ ~ \text{where} ~ x = 0, \, 1, \, 2, \, 3, \, 4, \, 5, \, 6$
The probability of getting 1 or more sixes when 6 dice are rolled
$= 1 - (_6C_0)(1/6)^0 (5/6)^{6}$
The probability of getting exactly x sixes when n dice are rolled
$= (_nC_x)(1/6)^x (5/6)^{n - x} ~ ~ ~ \text{where} ~ x = 0, \, 1, \, 2, \, \ldots , \, n$
The probability of getting n or more sixes when 6n dice are rolled is
$\begin{align} \displaystyle
P & = \sum_{x = n}^{6n} ({_{6n}C_x})(1/6)^x (5/6)^{6n-x} \\
& = 1 - \sum_{x = 0}^{n - 1} ({_{6n}C_x})(1/6)^x (5/6)^{6n - x}
\end{align}$
For 6 dice, n = 1. The probability of getting 1 or more sixes is
$\begin{align} \displaystyle
P_a & = \sum_{x = 1}^{6} ({_6C_x})(1/6)^x (5/6)^{6-x} \\
& = 0.665
\end{align}$
For 12 dice, n = 2. The probability of getting 2 or more sixes is
$\begin{align} \displaystyle
P_b & = 1 - \sum_{x = 0}^{1} ({_{12}C_x})(1/6)^x (5/6)^{12 - x} \\
& = 0.619
\end{align}$
For 18 dice, n = 3. The probability of getting 3 or more sixes is
$\begin{align}
P_c & = 1 - \sum_{x = 0}^{2} ({_{18}C_x})(1/6)^x (5/6)^{18 - x} \\
& = 0.597
\end{align}$
Pa > Pb > Pc
The most likely to occur is (a) ← Answer: [ A ]
See also this problem at Wolfram post