A block of steel (d=8.05g/cm3) weighs 157.4kg. How many whole solid cylindrical pegs with a radius of 0.500in. and a length of 2.00 in. can

Volume of one cylindrical peg = pi*r²*L  where r is the radius and L is the length.

Here we have r = 0.5 inches = 0.5*2.54 cm = 1.27 cm,  and L = 2 inches = 2*2.54 cm = 5.08 cm

The volume of one peg = pi*1.27²*5.08 cm³.

The mass of one peg is given by volume*density = pi*1.27²*5.08 cm³ * 8.05 g/cm³ = pi*1.27²*5.08*8.05 g

So the number of these you can get out of a block of mass 157.4kg is 154.7*1000/(pi*1.27²*5.08*8.05)

$${\frac{{\mathtt{157\,400}}}{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{1.27}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{5.08}}{\mathtt{\,\times\,}}{\mathtt{8.05}}\right)}} = {\mathtt{759.605\: \!010\: \!502\: \!234\: \!142\: \!2}}$$

Rounding down, this is 759 whole pegs (and this assumes we can reshape the block as necessary!).

Note that 157.4 kg is 157400 g, and I've converted to grams because the density is in terms of grams/cm³, not kg/cm³.