Answer
Fraction of $m_{In^{113}}=.0430$
Fraction of $m_{In^{115}}=.9570$
Work Step by Step
The fraction of naturally occurring $In^{113}$ will be denoted by $f$. The fraction of naturally occurring $In^{115}$ will be denoted by $t$. Since there are only two naturally occurring isotopes, $$f+t=1$$ Hence, $t=1-f$.
We calculate the weighted average via $$fm_{In^{113}}+(1-f)m_{In^{115}}=114.818$$
See problem 2.3 for the weighted average formula.
Rearranging, we have
$$f=\frac{114.818-m_{In^{115}}}{m_{In^{113}}-m_{In^{115}}}$$
Plugging in $m_{In^{113}}=112.904$ and $m_{In^{115}}=114.904$, we find $f=.0430$ and, hence, $t=1-f=.9570$
