Let \(A\) denote the event that a child has an allowance, and \(B\) that a child does chores.
The definition of conditional probability says that
\(\mathbb P(B|A)=\dfrac{\mathbb P(B\cap A)}{\mathbb P(A)}\)
Assuming the events are independent, you could expand the numerator as \(\mathbb P(B\cap A)=\mathbb P(B)\times\mathbb P(A)\). Otherwise, you would need to know that exact probability.
You're given that \(\mathbb P(A)=46\%\), while \(\mathbb P(B\cap A)=21\%\).
So, the desired conditional probability is
\(\mathbb P(B|A)=\dfrac{21\%}{46\%}\approx45.65\$\)
