Answer
Approximately 84.16667 ml of $HCl$.
Work Step by Step
$C(H^+)*V(H^+) = C(OH^-)*V(OH^-)$
Because each $Ba(OH)_2$ has 2 OH, the concentration of $OH^-$ will be the double of the $Ba(OH)_2$ concentration.
$[OH^-] = 2*[Ba(OH)_2] = 2 * 0.101M = 0.202M$
$[H^+] = 1*[HCl] = 1*0.12M = 0.12M$
$0.12 * V(H^+) = 0.202*50$
$V(H^+) = \frac{10.1}{0.12}$
$V(H^+) \approx 84.1667ml$