Let the depth of the cistern be h meters
Then 4.5\(\times\)3\(\times\)h = 50
So, h \(=\frac{50}{13.5}\)
h \(=\frac{100}{27}\)
Area of sheet required = lb + 2 (bh + lh)
= lb + 2h(l + b)
\(=[\frac{4.5\times3+2\times100}{27(4.5+3)}]\) sq m
\(=(\frac{13.5+200}{27\times7.5})\)
\(=(\frac{27}{2}+\frac{500}{9})\)
\(=\frac{1243}{18}\)
Therefore weight of lead \(=(\frac{27\times1243}{18})=\frac{3729}{2}\) kg
= 1864.5.
The correct option is (B)