Using the Law of Cosines

\cos A = \dfrac{b^2+c^2-a^2}{2bc}=\dfrac{2^2+3^2-4^2}{2\times 2\times 3 } =-\dfrac{1}{4}

According to Pythagorean Identity

\sin A = \sqrt{1-\cos^2 A} = \dfrac{\sqrt{15} }{4}

Using the Law of SInes

\dfrac{a}{\sin A}=2R

R = \dfrac{4}{2\times \dfrac{\sqrt{15} }{4} } = \dfrac{8}{\sqrt{15} }

Area formula of a triangle

A = \dfrac{1}{2}ab\sin C

According g to The Law of Sines,

\dfrac{c}{\sin C} = 2R

Then

Let s =\dfrac{a+b+c}{2} , then

s=\dfrac{2+3+4}{2} = \dfrac{9}{2}

According to Heron’s formula.

A = \sqrt{s(s-a)(s-b)(s-c)} ,

=\sqrt{\dfrac{9}{2}(\dfrac{9}{2}-2)(\dfrac{9}{2}-3)(\dfrac{9}{2}-4)}

=\dfrac{1}{4}\sqrt{9\cdotp 5\cdotp 3\cdotp 1 }

=\dfrac{3}{4}\sqrt{15}

Substituting to (1) yields

R = \dfrac{2\times 3\times 4 }{4\times \dfrac{3}{4}\sqrt{15} }

=8/\sqrt{15}