Answer
${\bf 1.19}\;\rm h$
Work Step by Step
We have a sample containing two types of radioactive atoms, $ A $ and $ B $. Initially:
There are five times as many $ A $ atoms as $ B $ atoms.
$$
(N_A)_0 = 5 (N_B)_0\tag 1
$$
After two hours, the number of $ A $ and $ B $ atoms become equal.
$$
N_A = N_B\tag 2
$$
And we are given that the half-life of $ A $ is 0.50 hours.
After $ t = 2 $ hours, the number of remaining atoms of $ A $ and $ B $is given by
$$
N_A = (N_A)_0 \left( \frac{1}{2} \right)^{t/(t_{1/2})_A}
$$
$$
N_B = (N_B)_0 \left( \frac{1}{2} \right)^{t/(t_{1/2})_B}
$$
From (2), these two equations are equal.
Thus,
$$
(N_A)_0 \left( \frac{1}{2} \right)^{t/(t_{1/2})_A} = (N_B)_0 \left( \frac{1}{2} \right)^{t/(t_{1/2})_B}
$$
Plug from (1);
$$
5 (\color{red}{\bf\not} N_B)_0 \left( \frac{1}{2} \right)^{t/(t_{1/2})_A} = (\color{red}{\bf\not} N_B)_0 \left( \frac{1}{2} \right)^{t/(t_{1/2})_B}
$$
After two hours, $t=2$ h, where $(t_{1/2})_A=0.5$ h
$$
5 \left( \frac{1}{2} \right)^{2/0.5} = \left( \frac{1}{2} \right)^{2/(t_{1/2})_B}
$$
$$ \frac{5}{16}= \left( \frac{1}{2} \right)^{2/(t_{1/2})_B}
$$
Take the Natural Logarithm of Both Sides:
$$
\ln\left(\frac{5}{16}\right) = \frac{2.0}{(t_{1/2})_B} \ln\left(\frac{1}{2}\right)
$$
Solve for $ t_{1/2}(B) $;
$$
(t_{1/2})_B = \frac{2.0}{ \ln\left(\frac{5}{16}\right) } \ln\left(\frac{1}{2}\right)
$$
$$
(t_{1/2})_B =\color{red}{\bf 1.19}\;\rm h
$$
