let f(x) = x^3-3x+k

so f^'(x) = 3 x^2 - 3

on comparing with 0, x = +1, -1

on double differentiating we get that function f(x) is minimum at +1 and maximum at -1

so for getting the only +ve real root plot a graph such that it could intersect positive x axis at any point and nowhere else. From the graph it is clear that k will be negative.

at x = +1,  f(x) = k-2

at x = -1, f(x) = k+2

so on the basis of given conditions k-2<0 or k<2  and  k+2<0 or k<-2 

finally k<-2 so [|k|] has minimum value 2