What volume of 6.00 m naoh would be required to increase the ph to 4.93?

According to this formula:

PH = Pka + ㏒[A^-]/[HA]

when we have the value of PH=4.93 & Pka (missing in your question) = 4.76 &

no. of mole of acetate = 40 mmol & no. of mole of acetic acid = 60

so by substitution:

4.93 = 4.76 + ㏒ ((40+X) / (60-X))

0.17 = ㏒ ((40+X) / (60-X))

∴X = 19.66 mmol = 0.019 mol

finally we can get the volume from this formula of the molarity:

molarity = number of moles / volume

6 = 0.019 / V

∴V = 0.0032 mL = 3.2 L