For sample 1: n1=12, m1=32750, s1=1900, variance, v1=s1^2=3610000
For sample 2: n2=10, m2=31200, s2=1825, v2=s2^2=3330625
Variance of source population=(v1+v2)/((n1-1)+(n2-1))=347031.25, because we're told that it is equal for both samples. We can now estimate the standard deviation S for the sampling distribution = √(V(1/n1+1/n2))=252.2348 approx.
The t-score=(m1-m2)/S=(32750-31200)/252.2348=6.145 approx. This tells us that the difference in means is more than 6 standard deviations.
The degrees of freedom are n1-1+n2-1=20. The value of over 6 well exceeds the confidence level of 99.9% (critical value for t=1.725 - 2-tail). So the claim is false: there is no significant difference between the income levels.