I think standard form is just AX^2 + BX + C
so the standard form of 64000 is
0X^2 + 0X + 64000
Not having your text or lecture notes I have no way of knowing what "standard form" means here. The most likely case is it is a number written with one digit before the decimal point multiplied by a power of .
So:
The last term on the right would in this case be standard form (you may or may not want to lose the trialling zeros depending on if they are significant or not)
In the second case you want to move the decimal point one place to the left which increases the power of by to give .
CB