Let the four numbers be 'a', 'b', 'c' and 'd'
Given ratio of first two numbers is 1:3 and four numbers are in proportion,
\((\frac{5-c}{3})^2+c^2=0\)
\(\frac{25+c^2-10c}{9}+c^2=5\)
\(25+c^2-10c+9c^2=5\)
\(10c^2-10c+20=0\)
\(c^2-c+10=0\)
After solving the equation we get, c = 2.
Substitute the value of 'c' in equation (2)
\(a=\frac{5-2}{3}\) (or) a=1
We know b = 3a (or) b = 3
We know d = 3c (or) d = 6
Average of the four numbers,
\(=\frac{1+3+2+6}{4}\)
=3
Hence, option B is the correct answer.The correct option is (B): 3