(A) 1 (B) 2 (C) 3 (D) None of the options given
Taking log, ${x^x}\ln |x| = \ln |x|$
$ \Rightarrow \ln |x|\left( {{x^x} - 1} \right) = 0$
$\ln |x| = 0,x = \pm 1$ Or ${x^x} = 1$
Taking log, $x\ln |x| = 0$
x = 0 Or $\ln |x| = 0,x = \pm 1$
x = 0 is rejected as it leads to $0^0$ situation.
Hence, two solutions or option (B).