Answer
The 3rd term is $\binom{7}{2}(-3)^2x^5=21\times 9 x^5=189x^5$
Work Step by Step
By using the binomial theorem we can expand the algebraic expression in the form of:
$(x+b)^n=\binom{n}{0}b^0x^n+\binom{n}{1}b^1x^{n-1}+...+\binom{n}{n-i}b^{n-i}x^{i}+...+\binom{n}{n-1}b^{n-1}x^1+\binom{n}{n}b^nx^0$
We expand the given expression:
$(x-3)^7=\binom{7}{0}(-3)^0x^7+\binom{7}{1}(-3)^1x^6+\binom{7}{2}(-3)^2x^5+\binom{7}{3}(-3)^3x^4+\binom{7}{4}(-3)^4x^3+\binom{7}{5}(-3)^5x^2+\binom{7}{6}(-3)^6x^1+\binom{7}{7}(-3)^7x^0$
The 3rd term is $\binom{7}{2}(-3)^2x^5=21\times 9 x^5=189x^5$
