How many grams is 4.2x1024 atoms of sulfur?

Molar mass of sulfur atom = 32.06 g/mol
Avogadro constant = 6.022 × 10²³ /mol

Mass of S atoms
= (4.2 × 10²⁴ S atoms) × (1 mol S / 6.022 × 10²³ S atoms) × (32.06 g S atom / 1 mol S atom)
= 224 g S atoms

Billrussell421 has given you the totally correct answer based on what you have submitted.
However I do not think that this is what you want to ask . Learn how to type exponents using the "power key" which is above the number 6 on the keyboard, above the alphabet letters. This is the ^ symbol
Now if you had asked: How many grams is 4.2x10^24 atoms of sulfur, you will get the answer:
Molar mass S = 32.066g/mol
1 mol S atoms = 6.022*10^23 S atoms
Therefore mass of 4.2*10^24 S atoms
Mass = 4.2*10^24 atoms / 6.022*10^23 atoms * 32.066g = 223.64g
Answer: 2 significant digits = 220g

One mole is 6.02 * 10^23 atoms. According to the periodic table, the mass of one mole of sulfur is 32.066 grams. Let’s use the following proportion to determine the mass.

32.066 ÷ 6.02 * 10^23 = x ÷ 4.2 * 10^24
x = 32.066 * 4.2 * 10^24 ÷ 6.02 * 10^23

Rounded to two significant digits, the mass of sulfur is approximately 220 grams.

4.2x1024 = 4300.8 atoms

6.022e23 molecules/mole Avogadro constant
sulfur mol weight = 32 g/mole

4300.8 atoms / 6.022e23 atoms/mole = 7.142e-21mole

32 g/mole x 7.142e-21mole = 2.29e-19 g

Remember
1 mole sulphur contains 32 g
Also
1 mole contains 6.022 x 10^23 atoms

Hence
4.2 x 10^24 / 6.022 x 10^23 = 6.976... moles.
Hence mass(S) = moles x Ar
mass(S) = 6.976... x 32 = 223.25 g