Answer
The fish must inflate the swim bladder to a volume of $1.7\times 10^{-7}~m^3$
Work Step by Step
We can find the volume $V$ of the fish with a deflated bladder:
$V = \frac{m}{\rho} = \frac{0.010~kg}{1080~kg/m^3} = 9.26\times 10^{-6}~m^3$
We can find the volume $V'$ which will give the fish a density of $1060~kg/m^3$:
$V' = \frac{m}{\rho} = \frac{0.010~kg}{1060~kg/m^3} = 9.43\times 10^{-6}~m^3$
The volume of the inflated swim bladder is equal to the difference of $V'$ and $V$:
$V'-V = (9.43\times 10^{-6}~m^3)-(9.26\times 10^{-6}~m^3)$
$V'-V = 1.7\times 10^{-7}~m^3$
The fish must inflate the swim bladder to a volume of $1.7\times 10^{-7}~m^3$.
