Specific activity just asks: for every microgram of enzyme protein you have, how many micromoles of substrate does it turn over each minute? Let's build that ratio directly from the given rate.
First convert the rate into the standard IU time base of per minute. The enzyme converts 10 mM substrate per hour. Since there are 60 minutes in an hour, the rate per minute is
\[ \frac{10\ \text{mM}}{60\ \text{min}} = 0.16667\ \text{mM min}^{-1} = 0.16667\ \text{mmol L}^{-1}\text{min}^{-1} \]Now switch this concentration based rate into an amount based rate by fixing a reaction volume of 1 mL, matching how the enzyme concentration, 10 $\mu$g mL$^{-1}$, is expressed. In 1 mL $= 0.001$ L, the amount of substrate converted per minute is
\[ 0.16667\ \text{mmol L}^{-1}\text{min}^{-1} \times 0.001\ \text{L} = 0.16667 \times 10^{-3}\ \text{mmol min}^{-1} = 0.16667\ \mu\text{mol min}^{-1} \]By definition, 1 IU is 1 $\mu$mol of substrate converted per minute, so this 1 mL of reaction holds 0.16667 IU of enzyme activity. That same 1 mL contains $10\ \mu\text{g}$ of enzyme protein, from the given 10 $\mu$g mL$^{-1}$.
Divide the activity by the mass of protein to get specific activity:
\[ \text{Specific activity} = \frac{0.16667\ \text{IU}}{10\ \mu\text{g}} = 0.016667\ \text{IU}\ \mu\text{g}^{-1} \approx 0.017\ \text{IU}\ \mu\text{g}^{-1} \]Let's summarize:
So the specific activity of the purified enzyme, rounded to three decimal places, is $0.017$ IU $\mu\text{g}^{-1}$.