1. We are given $f(x)=x^3+x^2 f'(1)+x f''(2)+f'''(3).$

Let $A=f'(1), \quad B=f''(2), \quad C=f'''(3).$ Then $f(x)=x^3+Ax^2+Bx+C.$

So $f(x)$ is a cubic polynomial.

2. Differentiate: $f'(x)=3x^2+2Ax+B,$ $f''(x)=6x+2A,$ $f'''(x)=6.$

Hence $C=f'''(3)=6.$

3. Now use the definitions of $A$ and $B$.

Since $A=f'(1)$, $A=3(1)^2+2A(1)+B=3+2A+B.$ So A=3+2A+B \implies A+B=-3. \tag{1}

Since $B=f''(2)$, B=6(2)+2A=12+2A. \tag{2}

Substitute (2) into (1): $A+(12+2A)=-3$ $3A+12=-3$ $3A=-15$ $A=-5.$

Then $B=12+2(-5)=2.$

4. Therefore $f'(x)=3x^2+2(-5)x+2=3x^2-10x+2.$

Now evaluate at $x=10$: $f'(10)=3(10)^2-10(10)+2=300-100+2=202.$

Therefore, the required integer is $\boxed{202}.$