You can write a formula for this:
\[P(t) = P_0(\frac{1}{2})^{(t/t_{\text{half}})}\]where \(P_0\) is the starting amount of the parent
\(P(t)\) is the amount of the parent remaining at time \(t\)
\(t_{\text{half}}\) is the half-life of the parent
If you try out some values:
\[P(0) = P_0(\frac{1}{2})^{0/t_{\text{half}}} = P_0(\frac{1}{2})^0 = P_0*1 = P_0\]At time \(t = 0\), we have all of the original parent
after 1 half-life (\(t = t_{\text{half}}\)) we have:
\[P(t_{\text{half}}) = P_0(\frac{1}{2})^{(t_{\text{half}}/t_{\text{half}})} = P_0(\frac{1}{2})^1 = \frac{P_0}{2}\]so as expected, 1 half-life means 1/2 of the parent is gone
after 2 half-lives:
\[P(2t_{\text{half}}) = P_0(\frac{1}{2})^{(2t_{\text{half}}/t_{\text{half}})} = P_0(\frac{1}{2})^2 = \frac{P_0}{4}\]
only 1/4 of the parent remains after 2 half-lives
All as we expect. We can use that formula to find the amount remaining at any time, not just multiples of the half-life:
\[P(6.64t_{\text{half}}) = P_0(\frac{1}{2})^{(6.64t_{\text{half}}/t_{\text{half}})} = P_0(\frac{1}{2})^{6.64} \approx \frac{P_0}{100}=0.01P_0\]after 6.64 half-lives, we have about 1% of the original remaining.
For the specific problem at hand, \(P_0 = 80\text{ grams}\) so
\[P(2\text{ half-lives} = (80\text{ grams})(\frac{1}{2})^{2/1} = \frac{80\text{ grams}}{4} = 20\text{ grams}\]
If we have \(80\) grams of parent material at the start, and only \(20\) grams remains, that means the daughter product is \(80 \text{ grams} - 20 \text{ grams} = 60 \text{ grams}\)
(This assumes that the daughter product is not ALSO decaying into something else in that same time, although the sum total of all the daughter products plus the parent would still add up to the starting amount)