x = $7^{-1}\ mod\ 216$
=> $(7 * x)\ mod\ 216 = 1$
We use the extended euclidean algorithm to compute x
$216 = 7 * 30 + 6\ (1)$
$7 = 6 * 1 + 1\ (2)$
In equation (2), we get 1 as remainder implying GCD (7,216) = 1 , therefore there exists a multiplicative inverse of 7 modulo 216
From equation (2)
$1 = 7 - 6 \ (3)$
From equation (1)
$6 = 216 - 7 * 30\ (4)$
Substituting 6 of equation (4) in equation (3)
$1 = 7 - (216 - 7 * 30)$
$=> 1 = 7 * 31 - 216$
Taking mod 216 on both sides
1 = (7*31) mod 216 - 216 mod 216
=> 1 = (7*31) mod 216
Thus x = 31