There's some very important information in the fine print.



FIRST SOLUTION
At 30 degrees Celsius, the solubility is about 40%.
This means the mass of lead nitrate comprises 40% of the entire solution
Let x = the number of grams of lead nitrate in the resulting solution
This means the mass of the entire solution = 10 + x grams
We can now write: x/(10 + x) = 40/100
Simplify: x/(10 + x) = 2/5
Cross multiply: (5)(x) = (2)(10 + x)
Expand: 5x = 20 + 2x
Solve: x = 20/3
In other words, the FIRST solution contains 20/3 grams of lead nitrate.


SECOND SOLUTION
At 60 degrees Celsius, the solubility is about 50%.
This means the mass of lead nitrate comprises 50% of the entire solution
Let y = the number of grams of lead nitrate in the resulting solution
We can write: y/(10 + y) = 50/100
Simplify: y/(10 + y) = 1/2
Cross multiply: (2)(y) = (1)(10 + x)
Expand: 2y = 10 + x
Solve: x = 10
In other words, the SECOND solution contains 10 grams of lead nitrate.


So the mass of lead nitrate went from 20/3 grams to 10 grams, and we need to find the percent increase.

WE COULD plug 20/3 and 10 into the formula for finding percent change, but that's going to be a pain.
Instead multiply both values by 3 to get the equivalent "ratios" 20 and 30

If we do this, we can see that there's a 50% increase from 20 and 30

Answer: 50

Aside: As @Carcass already noted, it's highly unlikely that this would be a Numeric Entry question, since we can't be very accurate when it comes to reading values off the chart.